I can't seem to prove that $e^a=\lim\limits_{x\rightarrow\infty}(1+\frac{a}{x})^x$. I'm sure there must be some algebraic manipulations that can be done in order to show that $\lim\limits_{x\rightarrow\infty}(1+\frac{a}{x})^x=$$\lim\limits_{x\rightarrow\infty}(1+\frac{1}{x})^{xa}$. What are these algebraic operations? I apologize in advance if maybe this question is a little too simple, but I can't for the life of me figure it out. Thanks.

$\lim_{x\to \infty} {\left({\frac{a}{x}+1}\right)}^x$ can also be written as $e^{\lim_{x \to \infty} x \log {(\frac{a}{x}+1)}}$ Now, the as the limit equals to $a$, the expression equals to $e^{a}$.
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gauravOct 11 '11 at 4:15

5

The proof depends on the precise definitions that one makes. For many introductions to calculus, you would start by taking the natural logarithm of $(1+a/x)^x$. But if $e$ is defined as the limit of $(1+1/y)^y$ as $y \to \infty$, you would, for positive $a$, note that $(1+a/x)^x=((1+a/x)^{x/a})^a$. As $x\to\infty$, $x/a \to \infty$, so $(1+a/x)^{x/a} \to e$.
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André NicolasOct 11 '11 at 4:20

In short: what definition of the exponential function are you using? Your limit is actually one definition used for the exponential function...
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Guess who it is.Oct 11 '11 at 4:29

@Hautdesert: In connection with some of the other comments here (e.g. it depends on how $e$ is defined), some or all of the various manipulations that I recently posted in the ap-calculus listserv (use the URL that follows) could be of use in your class. mathforum.org/kb/message.jspa?messageID=7575015
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Dave L. RenfroOct 11 '11 at 14:41

For $a>0$, you can replace $x$ with $ax$ in the expression: as $x\to\infty$, so does $ax$.

For $a=0$, the limit is trivial.

For $a<0$, you can flip the expression upside down as $$\large\frac{1}{\left(1-\frac{a}{x+a}\right)^x},$$ then replace $x$ with $-a(x+1)$ (it goes to $\infty$ just the same as well): $$\large\frac{1}{\left(1+\frac{1}{x}\right)^{-a(x+1)}} .$$

As André Nicolas comments, often this equation is used to define $e$, for which case he has given a proof. Assuming instead that $e^x := \lim_{n\to\infty}\sum_{i=0}^{n} \frac{x^i}{i!}$, expand the binomial and show that the corresponding terms have equal coefficients.